Unfused spur.

I have now read you many times on this subject Andy on various posts. The point is that you are not understanding this at all. Let's look at the TN conditions again if you like.  In the most simplistic terms, say the CPC was a fuse, what time would be required to cause it to fail at the available fault current? Let's look at the available fault current first, what is it? At the DB, Z says it is about 890A. We have about 15 m doubled in the ring, so 7.5m The resistance of this is about 135 mOhms. The mains supply impedance is 0.26 Ohms by calculation. Therefore the L-N PSCC is 230/0.397 = 578A (About Mikes figure). At this current the breaker opens in less than 10ms, giving a maximum I²t of 3341 A²s. What is the problem now?  It is even lower for a L-E fault because the Earth conductor has higher resistance. This is pretty simple electrical design stuff. You are imagining something which is not correct or possible. Z is perfectly safe. In this case, we will not even get to 115°C, the design criterion for multiple damage-free faults, which strictly speaking is not mandated by BS7671. Now the TT version, the fault current must pass through the Re, so is very much smaller, the RCD is subjected to a very small I²t, as is the 1mm conductor.


(Joke) Here endeth the first lesson!

I have now read you many times on this subject Andy on various posts. The point is that you are not understanding this at all. Let's look at the TN conditions again if you like.  In the most simplistic terms, say the CPC was a fuse, what time would be required to cause it to fail at the available fault current? Let's look at the available fault current first, what is it? At the DB, Z says it is about 890A. We have about 15 m doubled in the ring, so 7.5m The resistance of this is about 135 mOhms. The mains supply impedance is 0.26 Ohms by calculation. Therefore the L-N PSCC is 230/0.397 = 578A (About Mikes figure). At this current the breaker opens in less than 10ms, giving a maximum I²t of 3341 A²s. What is the problem now?  It is even lower for a L-E fault because the Earth conductor has higher resistance. This is pretty simple electrical design stuff. You are imagining something which is not correct or possible. Z is perfectly safe. In this case, we will not even get to 115°C, the design criterion for multiple damage-free faults, which strictly speaking is not mandated by BS7671. Now the TT version, the fault current must pass through the Re, so is very much smaller, the RCD is subjected to a very small I²t, as is the 1mm conductor.


(Joke) Here endeth the first lesson!